Question

A point charge q₁ = + 7.00 x 10^-10 C is at the point x = 0.900m, y= 0.700 m and a second point charge q₂ = -3.00 x 10^-10 C is at the point x = 0.900 m, y= 0m . Calculate the magnitude and direction of the resultant electric field at the origin due to these charges. 

Answer 1

Electric field due to charge$q_1$

$E_1=\frac{kq_1}{r^2}$

$E_1=\frac{9\times10^9\times7\times10^{-10}}{\sqrt{(1.3})^2}=4.846N/C$

$E_2=\frac{kq_2}{r^2}$

$$E_2=-\frac{9\times10^9\times3\times10^{-10}}{{(0.9})^2}=-3.33N/C$$

$E_{net}=\sqrt{E_1^2+E_2^2+2E_1E_2cos\phi}$

$\phi=45°$

$$E_{net}=\sqrt{4.846^2+(-3.33)^2+2\times4.846\times3.33\times cos135}$$

$E_{net}=3.42N/C$

Direction OP direction according to daigram